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Math Help - Derivative help

  1. #1
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    Derivative help

    Question:

    f(x) = ln(cosx)

    I need to find the derivative.

    I'm unsure where to start. I tried the chain rule, and I get stuck, as follows:

    \frac {1}{x} (cosx)^0 (-sinx)

    \frac {-sinx}{x}

    The answer is supposed to be -tanx
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  2. #2
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    Quote Originally Posted by Glitch View Post
    f(x) = ln(cosx)

    I need to find the derivative.
    make u = \cos x \implies y = \ln u

    \frac{du}{dx} = -\sin x and \frac{dy}{du}= \frac{1}{u}

     <br />
\frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx}= \frac{1}{u}\times -\sin x = \frac{-\sin x}{\cos x} = -\tan x <br />
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  3. #3
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    Interesting. So that's the chain rule? Your working out isn't very familiar to me.
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  4. #4
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    Quote Originally Posted by Glitch View Post
    Interesting. So that's the chain rule? Your working out isn't very familiar to me.
    hi
    if u have two functions of x, f and g,how do you find the derivative of f(g(x)) ?
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  5. #5
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    I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.
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  6. #6
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    Quote Originally Posted by Glitch View Post
    I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.
    well ..
    \left (\ln f(x)  \right )'=\frac{f'(x)}{f(x)}
    \left (\ln \cos(x)  \right )'=\frac{-\sin(x)}{\cos(x)}
    (\cos(x))'=-\sin(x)
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  7. #7
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    Yeah, I made an error. Thanks for the help!
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  8. #8
    MHF Contributor chisigma's Avatar
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    A general formula for the 'logarithmic derivative' of a function is...

    d/dx ln f(x) = f'(x)/f(x) (1)

    ... and here is f(x) = cos x , so that...

    Kind regards

    \chi \sigma
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  9. #9
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    Quote Originally Posted by Glitch View Post
    I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.
    Why were you lowering the power, though? The derivative of ln{x} is 1/{x} and the derivative of cosx is -sinx. That should be sufficient:

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